题目描述
上图是一个电话的九宫格，如你所见一个数字对应一些字母，因此在国外企业喜欢把电话号码设计成与自己公司名字相对应。例如公司的Help Desk号码是4357，，因为4对应H、3对应E、5对应L、7对应P，因此4357就是HELP。同理，TUT-GLOP就代表888-4567、310-GINO代表310-4466。NowCoder刚进入外企，并不习惯这样的命名方式，现在给你一串电话号码列表，请你帮他转换成数字形式的号码，并去除重复的部分。输入描述:输入包含多组数据。每组数据第一行包含一个正整数n（1≤n≤1024）。紧接着n行，每行包含一个电话号码，电话号码仅由连字符“-”、数字和大写字母组成。没有连续出现的连字符，并且排除连字符后长度始终为7（美国电话号码只有7位）。输出描述:对应每一组输入，按照字典顺序输出不重复的标准数字形式电话号码，即“xxx-xxxx”形式。每个电话号码占一行，每组数据之后输出一个空行作为间隔符。输入例子:124873279ITS-EASY888-45673-10-10-10888-GLOPTUT-GLOP967-11-11310-GINOF101010888-1200-4-8-7-3-2-7-9-487-32794UTT-HELPTUT-GLOP310-GINO000-1213输出例子:310-1010310-4466487-3279888-1200888-4567967-1111000-1213310-4466888-4357888-4567
// write your code here cpp#include <iostream>#include <string.h>#include <map>using namespace std;bool IsOk(char *str){int count = 0;int n = strlen(str);char *p = str;while (*p != '\0'){if (*p == '-')count++;p++;}if (n – count != 7)return false;return true;}bool isNum(char ch){return (ch – '0') >= 0 && (ch – '0') <= 9;}int main(){int n;char inputStr[1024]; char StrBuff[256] = {0};StrBuff['A'] = '2';StrBuff['B'] = '2';StrBuff['C'] = '2';StrBuff['D'] = '3';StrBuff['E'] = '3';StrBuff['F'] = '3';StrBuff['G'] = '4';StrBuff['H'] = '4';StrBuff['I'] = '4';StrBuff['J'] = '5';StrBuff['K'] = '5';StrBuff['L'] = '5';StrBuff['M'] = '6';StrBuff['N'] = '6';StrBuff['O'] = '6';StrBuff['P'] = '7';StrBuff['Q'] = '7';StrBuff['R'] = '7';StrBuff['S'] = '7';StrBuff['T'] = '8';StrBuff['U'] = '8';StrBuff['V'] = '8';StrBuff['W'] = '9';StrBuff['X'] = '9'; StrBuff['Y'] = '9';StrBuff['Z'] = '9';StrBuff['-'] = '-';//cin >> n;while (cin>>n){map<string, int> mp;while (n–){cin >> inputStr;string s;if (IsOk(inputStr)){char *p = inputStr;while (*p != '\0'){if (s.size() == 3){s += '-';continue;}else{if (isNum(*p)){s += *p;}else{if (*p != '-')s += StrBuff[*p];}}p++;}mp.insert(pair<string, int>(s, 0));}}map<string, int> ::iterator it = mp.begin();while (it != mp.end()){cout << (*it).first.c_str() << endl;it++;}cout << endl;}return 0;}
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一张单程车票，一颗潇洒的心。
